Advent of Code 2016 January 4, 2017
I've been doing the Advent of Code 2016 this year, I've been a little late to the game due to a week off of doing anything with computers and generally just not being very attentive to it.
There have been a few favorites. Day 11, 13 and 17 with their tree searching have been a good refresher. Day 15 literally took me longer to understand the problem than to code it up.
Day 19 took me through a few different paths. Day 19 is the White Elephant with 3 million elves. The elf takes the presents from the elf to their left and the next elf that has presents goes next. And it repeats until one elf has all the presents. Part 2, which you find out later, is that it changes it to take the presents from the elf across from them.
The first attempt at this solution was an array of elves. 3 million odd elves. Looping through would be fast but finding the next elf who has presents would be looping a ton. Solving the 5 elf example worked but moving it to 3 million+ pointed out how terribly inefficient this would be.
So. Linked list to the rescue (I have always said "to the rescue" when picking a tool to solve the job, but ever since seeing "The Martian", now my internal voice is replaced by Matt Damon's, when he determines that hexadecimal is the answer to his problems).
For part one, just create 3 million elves and have a pointer to the next elf. The last elf's next elf is the first elf. Done.
for elf.Next != elf
elf.Presents += elf.Next.Presents
elf.Next = elf.Next.Next
elf = elf.Next // move to the next elf
This solves it very quickly. But then the bombshell of part 2 comes in. Find the elf across from the current elf in a Linked List?! What?
First attempt was for each elf, use a counter to determine the one across. So every time an elf is removed from the circle, decrement the counter. To determine the across elf, loop to across.Next (count / 2) times. This works. But then it proved very slow as well. Each time you're looping it loops 1.5 million-ish times (half). This proved to be correct though, with my test with the five elves.
Then I remembered the ability to just have as many pointers as you want! This is computers, you can do whatever you want. There are no rules :) So, instead I created an "across" pointer and initialized it with my count/2 loop. This would be the only time it would need to loop. Just determine it's starting location.
I would still have to keep a count though. The count was useful for the case where, for instance in the 5 elf circle... 1 takes presents from 3, but there are 2 across from 1, so you choose the left side. However, removing 3 then, and moving to 2, 2 is across from 5, but if we were to just increment the across elf to across.Next, it would point to 4. So if there are an even number of elves left, add another move to across.Next after setting it to across.Next (move it two forward in the list). This proved to work and was very fast :)
Check the solution here - https://github.com/jasontconnell/advent/blob/develop/2016/19.go